Let $F$ be an intersection point of altitude $CD$ and internal angle bisector $AE$ of right angled triangle $ABC$, $\angle ACB = 90^{\circ}$. Let $G$ be an intersection point of lines $ED$ and $BF$. Prove that area of quadrilateral $CEFG$ is equal to area of triangle $BDG$
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2015
Tags: geometry, angle bisector
dikhendzab
12.05.2020 21:11
Since $AE$ is bisector we have that $\frac{AC}{AB}=\frac{CE}{BE}$ and $\frac{AD}{AC}=\frac{DF}{CF}$. From similarity of triangles $ABC$ and $ADC$ we have that $\frac{AD}{AC}=\frac{AC}{AB} \implies \frac{CE}{BE}=\frac{DF}{CF} \implies CE \cdot CF=BE \cdot DF=(BC-CE)(CD-CF)$ $CE \cdot CF=BC \cdot CD-BC \cdot FC-CE \cdot CD+CE \cdot CF \implies BC \cdot CD=BC \cdot CF+CD \cdot EC$ $A_{BCD}=\frac{1}{2}BC \cdot CD \cdot \sin \angle BCD=\frac{1}{2}BC \cdot CF \cdot \sin \angle BCD+\frac{1}{2}EC \cdot CD \cdot \sin \angle BCD$ So, $A_{BCD}=A_{BCF}+A_{ECD}$ From obvious reasons $A_{BCD}=A_{ECD}+A_{BEG}+A_{BGD}=A_{BCF}+A_{ECD}$ and $A_{BCD}=A_{BCF}+A_{BGD}+A_{GDF}=A_{BCF}+A_{ECD}$ We will now get $A_{BEG}+A_{BGD}=A_{BCF}$ and $A_{BGD}+A_{GDF}=A_{ECD}$. We will add up these equations, so: $2A_{BGD}+A_{BEG}+A_{GDF}=A_{BCF}+A_{ECD}$ $2A_{BGD}=A_{BCF}+A_{ECD}-A_{GDF}-A_{BEG}=2A_{CGFE}+A_{GDF}-A_{GDF}+A_{BEG}-A_{BEG}=2A_{CGFE}$. Finally: $A_{BGD}=A_{CGFE}$
dgreenb801
13.05.2020 09:16
Let $H$ be the foot of the perpendicular from $E$ to $AB$, then $CE=HE$ as $\triangle ACE$ and $\triangle AHE$ are congruent. Note that $\angle CFE = \angle AFD= 90 - \angle FAD = 90 - \angle FAC = \angle FEC$. Thus, $CF=CE=HE$. Letting $[ABC]$ denote the area of $\triangle ABC$, we have $[BED] = \frac{1}{2} HE \cdot BD = \frac{1}{2} CF \cdot BD = [BFC]$. Thus, $[BED]-[BEG]=[BFC]-[BEG]$, or $[BDG]=[CEFG]$.